129.
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1) ( 0,1) ^ (- Lg 0,3 ) =(10⁻¹ )* ^ (- Lg 0,3 ) =(10)* ^ (Lg 0,3 <span>) = 0,3 .
2) 10 ^(-Lg4) = </span>10 ^(Lg4 ⁻¹ ) = 4 ⁻<span>¹ = 1/4. || 0,25 ||
3) 5 ^( -Log 5 3) = </span> 5 ^( Log 5 3 ⁻¹ ) = 3 ⁻<span>¹ =1/3.
4) 1/ 6 ^(-Log 6 4) =( 6</span>⁻¹)^ (-Log 6 4) =6^ (Log 6 4) ) <span> =4.
130.
</span>------<span>
1) 4Log(1/2) 3 -(2/3) Log (</span>1/2) 27 -2 Log (<span>1/2) 6 =
</span>Log(1/2) 3⁴ - Log (1/2) (3³)^(2/3) - Log (1/2) 6² =
Log(1/2) 3⁴ - Log (1/2) 3² - Log (1/2) 6² = Log(1/2) 3⁴/(3²* 6<span>²) =
</span>Log(1/2) 3²/ 6² = Log(1/2) (3/ 6)² = Log(1/2) (1/ 2)<span>² <span>=2.
</span>или по другому:
</span>4Log(1/2) 3 -(2/3) Log (1/2) 27 -2 Log (<span>1/2) 6 =
</span>4Log(1/2) 3 - (2/3)Log( 1/2) 3³ -2 Log (1/2) 6 =
4Log(1/2) 3 - 2 Log (1/2) 3 - 2Log (1/2) 6 =2Log(1/2) 3 - 2 Log (1/2) 6=
2*(Log(1/2) 3 - Log (1/2) 6 ) =2 Log (1/2) (3/6 )=2 <span>Log (1/2) (1/2 ) =2</span>
Log(1/2) 3²/ 6² = Log(1/2) (3/ 6)² = Log(1/2) (1/ 2)<span>² <span> =2.</span></span>
2) ( 2/3)*Lg0,001 +Lg ∛1000 -(3/5)*Lg √10000 =
( 2/3)*Lg10⁻³ +Lg ∛10³ -(3/5)*Lg √10⁴ = ( 2/3)*(-3) +Lg10 -(3/5)*Lg 10² = <span>
(2/3)*(-3) +1 -(3/5)*2 = -2 +1 - 1,2 = - 2,2 .</span>
1)tg(π/4)+ctg(5π/4) = tg(π/4)+ctg(π+π/4) = tg(π/4)+ctg(π/4) = 1+1 = 2.
2)ctg(π/3)-tg(π/6) = 1/3^0.5-1/3^0.5 = 0.
*)3^0.5 - 'квадратный корень из трех'.
По всей видимости условие таково, что в знаменателе стоит выражение x^2-6x-27, тогда нахождение области определения сводится к тому, что знаменатель не равен нулю( на нуль делить нельзя!!!)